First of all I am sorry because I have asked similar kind of question a few days ago.But I still have problem with row reductions when there are letters in a matrix.The question is asking the value of 'a' when the rank of matrix is 1 , 2 , 3 and 4. I am not good at row reductions.In each row operations the matrix became more confusing.If someone help me with the row reductions , I will be very happy.
$$
\left(
\begin{array}{ccc}
1&1&2&0\\
2&a+1&3&a-1\\
-3&a-2&a-5&a+1\\
a+2&2&a+4&-2a
\end{array}
\right)
$$

Why are you reverting the linear-algebra tag? It's been added twice by two different people, you even responded positively to the comment suggesting that it be added; why have you twice removed it?
–
jorikiJun 21 '11 at 21:31

Not the title, the tags at the bottom.
–
Arturo MagidinJun 21 '11 at 21:36

We are just doing algebra, only in several columns at the same time. Keep going until you get a matrix for which the rank will be easy to figure out, depending on the values of $a$. I've gotten you almost all of the way there.

if rank=1, that means all the rows are dependent. The first row is given, so every other row should be a multiple of the first row. Let's start with the second row. since the first element on the second row is 2, and the first element on the first row is 1, then (a+1) should also be 2*1 and (a-1) should be 2*0. so $a$ should be equal to 1. However if you check the third row, substituting $a$ with 1, we have [-3, -1, -4, 2] which is not a multiple of the first row. hence rank=1 has no solution.

You can do row manipulation to make the matrix lower triangular. then the determinant would be the product of the elements on the diagonal.